# HELP WITH DERIVATIVE PROBLEM PLEASE:)?

+1 vote
DERIVE y=2ln(2x)+x
asked Jan 3, 2013 in CALCULUS

+1 vote

The given eqution is y=2ln(2x)+x

derivative on each side with respect to x

dy/dx = d/dx (2ln(2x)+x)

= d/dx (2ln(2x)) + d/dx (x)

formula d/dx (lnx) = 1/x and d/dx (x) = 1

dy/dx = 2 (1/2x) + 1

cancel comman terms

dy/dx = (1/x) + 1

The derivative of = 2ln(2x ) + x  is 2/x  + 1.

+1 vote

DERIVE y=2ln(2x)+x

Given that y = 2 ln(2x) + x

Derivative with respect to x

dy / dx = 2 d (ln(2x)) / dx + d(x) / dx

dy / dx = 2 [1 / 2x ] d(2x) / dx + 1              (d (ln(2x)) / dx = 1 / 2x )

dy / dx =  [1 / x ] d(2x) / dx + 1

dy / dx =  [1 / x ] 2 + 1                                (d(2x) / dx = 2)

dy / dx =  [2 / x ] + 1

The derivative of the y = 2 ln(2x) + x is [2 / x ] + 1

+1 vote
Given equation is  y=2ln(2x)+x

Derivative each with respect to " x "

dy /dx  = d / dx (2 ( ln (2x) ) +x  )

dy /dx  = 2 ( d/dx ( ln(2x) ) +d /dx (x)          (d/dx ln x = 1 / x; and d/dx(x) = 1)

dy /dx   = 2 (1 / 2x d/dx (2x) )+1               (d / dx (kx) = k)

dy /dx  = 2 (1 / 2x ) (2)  +1

dy /dx  = 4 / 2x  + 1

dy /dx    = 2 / x  + 1